Finding the $n$th term for the sequence $1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \dots$

Question

I have tried using a negative exponent. I need one statement not two, the pattern is $$1, \; \frac{1}{2}, \; 3, \; \frac{1}{4}, \; 5, \; \frac{1}{6}, \dots$$

Answer 1

$$1, \frac{1}{2} , 3, \frac{1}{4} , 5, \frac{1}{6} \dots$$

The nth term would be: $n$ if $n$ is odd, and $\frac{1}{n}$ if $n$ is even.

Answer 2

Hint: You can express it by $n^{a_n}$, where the $a_n$ give the appropriate power needed. For $n$ even, you want $a_n=1$ and for $n$ odd, you want $n=-1$ (recall $x^{-y}=1/x^y$).

So what formula gives $1,-1,1,-1,\ldots$?

Answer 3

The answer is simple! $(2n+1),\frac{n}{2}$ where $n$EZ.